Question

If Ais any square matrix, then matrix $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathit{A}\mathbf{-}{\mathit{A}}^{\mathbf{T}}\mathbf{)}$ is skew-symmetric.

Short answer

True.

Step-by-step solution

Step by step solution Step 1: Skew-symmetric matrices

Consider an m×nmatrix A.

The transpose${\mathit{A}}^{\mathbf{T}}$of Ais the$\mathit{n}\mathbf{\times }\mathit{m}$matrix whose$\mathit{i}{\mathit{j}}^{\mathbf{t}\mathbf{h}}$entry is$\mathit{j}{\mathit{i}}^{\mathbf{t}\mathbf{h}}$the entry of A: The roles of rows and columns are reversed.

We say that a square matrix Ais symmetricif ${\mathit{A}}^{\mathbf{T}}\mathbf{=}\mathit{A}$, and Ais called skew-symmetricif ${\mathit{A}}^{\mathbf{T}}\mathbf{=}\mathit{A}$.

Determine whether the statement is true or false

Consider the equation below.

$\begin{array}{rcl}(1/2(A-A^T\left)\right)& =& 1/2(A^T-A)\\ & =& -1/2(A-A^T)\end{array}$

Thus, by the definition of skew symmetric matrices statement is to be true.

Hence, the statement is true.